# Integrate $1/x$ in the sense of distribution, with non-symmetric excluded neighborhood of zero

I'm kind of confused with this question because of the boundary terms. for a test function $\phi$ define: $u(\phi) = \lim_{\epsilon \rightarrow 0+} \int_{x\notin(-3\epsilon,5\epsilon)} \phi(x) x^{-1} dx$

Find a locally integrable function $v$ such that $v'=u$ in the sense of distributions.

How can I deal with boundary terms to find the correct distribution?

• Since $u$ is the distribution defined (roughly) by integration against $1/x$, then we should expect the "integral" to be something like $\log|x|$. Try showing that the derivative of $\log|x|$ in the sense of distributions is $u$! – yousuf soliman Jul 21 '17 at 4:18
• well, I actually know that it will involve $log x$ , but I don`t see how the boundary will change anything. Should I for example divide the integral based on the interval and then deal with that? – Danny Jul 21 '17 at 15:14
• It will be $\log|x|$, not $\log(x)$. The limit is there to make sure everything is well defined. The integral $\int_{\mathbb{R}}\frac{\phi(x)}{x}~dx$ need not exist / make sense in the usual sense of the Lebesgue integral. You should write your integral as $\int_{-\infty}^{-3\epsilon} + \int_{5\epsilon}^{\infty}$ and then just apply integration by parts. Finally, be careful about your passage to the limit. – yousuf soliman Jul 21 '17 at 15:20
• @yousufsoliman It's not exactly $\log |x|$; the asymmetry of the excluded neighborhood of $0$ contributes something substantial. – user357151 Jul 21 '17 at 23:22
• Ahh my bad. I didn't work through any of the details, but your answer looks good! – yousuf soliman Jul 21 '17 at 23:24

The goal is to find a locally integrable $v$ such that $$\lim_{\epsilon \to 0+} \int_{x\notin(-3\epsilon,5\epsilon)} \phi(x) x^{-1} dx = -\int_{\mathbb{R}} \phi'(x)v(x)\,dx$$ As yousuf soliman suggested, use integration by parts, where the boundary term at $\infty$ does not appear because $\phi$ has compact support: $$\int_{5\epsilon}^\infty \phi(x) x^{-1} dx = -\phi(5\epsilon) \log(5\epsilon) - \int_{5\epsilon}^\infty \phi'(x) \log x\, dx$$ and $$\int_{-\infty}^{-3\epsilon} \phi(x) x^{-1} dx = \phi(-3\epsilon) \log(3\epsilon) - \int_{-\infty}^{-3\epsilon} \phi'(x) \log x\, dx$$ Add and let $\epsilon\to 0$. The integrals are okay because $\log |x|$ is integrable near $0$, but the boundary term requires some care. Subtracting off $\phi(0)$ helps here:
$$\phi(-3\epsilon) \log(3\epsilon) -\phi(5\epsilon) \log(5\epsilon) = (\phi(-3\epsilon)-\phi(0)) \log(3\epsilon) - (\phi(5\epsilon) -\phi(0) \log(5\epsilon) + \phi(0) \log (3/5)$$ The first two terms are $O(\epsilon \log \epsilon)$, so they tend to $0$. The limit is $\phi(0) \log (3/5)$. Summary of what we have so far:
$$u = \log(3/5) \delta_0 + (\log |x|)'$$ It remains to recall that the Dirac function $\delta_0$ is the distributional derivative of the Heaviside function $H$. $$v(x) = \log (3/5) H(x) + \log |x|$$ (up to an additive constant, of course)

• thanks a lot for you, It was very confusing for me. – Danny Jul 22 '17 at 23:47